## UGC NET CS 2015 Dec- paper-2

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Question 1 |

How many committees of five people can be chosen from 20 men and 12 women such that each committee contains at least three women?

75240 | |

52492 | |

41800 | |

9900 |

Question 1 Explanation:

Given data,

-- 20 men and 12 women

-- 5 people can choose from men and women

-- Each committee contains at least three women

Step-1: We have a constraint that each committee contains atleast 3 women.

possibility-1: 2 men + 3 women.

possibility-2: 1 men + 4 women.

possibility-3: 0 men + 5 women.

Step-1: They are asking to find all possibilities.

= (possibility -1)+ (possibility -2) + (possibility -3)

= . (

= (190*220) + (20*495) + (1*792)

= 41800 + 9900 + 792

= 52492

-- 20 men and 12 women

-- 5 people can choose from men and women

-- Each committee contains at least three women

Step-1: We have a constraint that each committee contains atleast 3 women.

possibility-1: 2 men + 3 women.

possibility-2: 1 men + 4 women.

possibility-3: 0 men + 5 women.

Step-1: They are asking to find all possibilities.

= (possibility -1)+ (possibility -2) + (possibility -3)

= . (

^{20} C_{ 2}. *^{12} C_{ 3} ) + (^{20} C_{ 2}. *^{12} C_{2} ) + (^{20} C_{ 2}. *^{ 12} C_{ 2} )= (190*220) + (20*495) + (1*792)

= 41800 + 9900 + 792

= 52492

Question 2 |

Which of the following statement(s) is/are false ?

(a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.

(b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.

(c) A complete graph (K

(d) A cycle over six vertices (C

(a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.

(b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.

(c) A complete graph (K

_{ n} ) has a Hamilton Circuit whenever n ≥ 3.(d) A cycle over six vertices (C

_{6} ) is not a bipartite graph but a complete graph over 3 vertices is bipartite.(a) only | |

(b) and (c) | |

(c) only | |

(d) only |

Question 2 Explanation:

(a)TRUE: A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.

(b)TRUE: A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.

(c)TRUE: A complete graph (K

(b)TRUE: A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.

(c)TRUE: A complete graph (K

_{n} ) has a Hamilton Circuit whenever n ≥ 3. (d) FALSE: A cycle over six vertices (C_{ 6} ) is not a bipartite graph but a complete graph over 3 vertices is bipartite.Question 3 |

Which of the following is/are not true?

(a) The set of negative integers is countable.

(b) The set of integers that are multiples of 7 is countable.

(c)The set of even integers is countable.

(d)The set of real numbers between 0 and 1⁄2 is countable.

(a) The set of negative integers is countable.

(b) The set of integers that are multiples of 7 is countable.

(c)The set of even integers is countable.

(d)The set of real numbers between 0 and 1⁄2 is countable.

(a) and (c) | |

(b) and (d) | |

(b) only | |

(d) only |

Question 3 Explanation:

(a)TRUE: The set of negative integers is countable.

Suppose negative integers set size is 10.

Ex: -1, -2, -3,.....,-10 is countable

(b)TRUE: The set of integers that are multiples of 7 is countable.

Suppose set of integers size is 10.

Ex: 1*7, 2*7, 3*7, .....,10*7 is countable

(c)TRUE: The set of even integers is countable.

Suppose set of even integers size is 10.

Ex: 2,4,6,8,10,....,20

(d) FALSE: The set of real numbers between 0 and 1⁄2 is countable. We can’t count real numbers.

Ex: 0.1, 0.2, 0.3, .....,0.∞

Suppose negative integers set size is 10.

Ex: -1, -2, -3,.....,-10 is countable

(b)TRUE: The set of integers that are multiples of 7 is countable.

Suppose set of integers size is 10.

Ex: 1*7, 2*7, 3*7, .....,10*7 is countable

(c)TRUE: The set of even integers is countable.

Suppose set of even integers size is 10.

Ex: 2,4,6,8,10,....,20

(d) FALSE: The set of real numbers between 0 and 1⁄2 is countable. We can’t count real numbers.

Ex: 0.1, 0.2, 0.3, .....,0.∞

Question 4 |

Consider the graph given below: The two distinct sets of vertices, which make the graph bipartite are:

(v _{1} , v_{ 4} , v_{ 6} ); (v_{ 2} , v_{ }3 , v_{ 5} , v_{ 7} , v_{ 8} ) | |

(v _{1} , v_{ 7} , v_{ 8} ); (v_{ 2} , v_{ 3} , v_{ 5} , v_{ 6} ) | |

(v _{1} , v_{ 4} , v_{ 6} , v_{ 7} ); (v_{ 2} , v_{ 3} , v_{ 5} , v_{ 8} ) | |

(v _{1} , v_{ 4} , v_{ 6} , v_{ 7} , v_{ 8} ); (v_{ 2} , v_{ 3} , v_{ 5} ) |

Question 4 Explanation:

A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets U and V such that every edge connects a vertex in U to one in V. Vertex sets U and V are usually called the parts of the graph. Equivalently, a bipartite graph is a graph that does not contain any odd-length cycles.

→ The two sets U and V may be thought of as a coloring of the graph with two colors.

Option A: FALSE because V

Option-B FALSE because V

Option-C TRUE because it follows properties of bipartied and no two colours are adjacent.

Option-D FALSE because because V

→ The two sets U and V may be thought of as a coloring of the graph with two colors.

Option A: FALSE because V

_{5} , V_{ 7} and V_{ 3} are adjacent. So, it not not bipartite graph.Option-B FALSE because V

_{5} , V_{ 6} and V_{ 2} are adjacent. So, it not not bipartite graph.Option-C TRUE because it follows properties of bipartied and no two colours are adjacent.

Option-D FALSE because because V

_{4} , V_{ 6} and V_{ 8} are adjacent. So, it not not bipartite graph.Question 5 |

A tree with n vertices is called graceful, if its vertices can be labelled with integers 1, 2,....n such that the absolute value of the difference of the labels of adjacent vertices are all different. Which of the following trees are graceful?

(a)

(b)

(c)

(a)

(b)

(c)

(a) and (b) | |

(b) and (c) | |

(a) and (c) | |

(a), (b) and (c) |

Question 5 Explanation:

Above all graphs are graceful.

Question 6 |

Which of the following arguments are not valid ?

(a) “If Gora gets the job and works hard, then he will be promoted. If Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard”.

(b) “Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty”.

(c) If n is a real number such that n >1, then n

(a) “If Gora gets the job and works hard, then he will be promoted. If Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard”.

(b) “Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty”.

(c) If n is a real number such that n >1, then n

^{ 2} >1. Suppose that n^{ 2} >1, then n >1.(a) and (c) | |

(b) and (c) | |

(a), (b) and (c) | |

(a) and (b) |

Question 7 |

Let P(m, n) be the statement “m divides n” where the Universe of discourse for both the variables is the set of positive integers. Determine the truth values of the following propositions.

(a)∃m ∀n P(m, n)

(b)∀n P(1, n)

(c) ∀m ∀n P(m, n)

(a)∃m ∀n P(m, n)

(b)∀n P(1, n)

(c) ∀m ∀n P(m, n)

(a) - True; (b) - True; (c) - False | |

(a) - True; (b) - False; (c) - False | |

(a) - False; (b) - False; (c) - False | |

(a) - True; (b) - True; (c) - True |

Question 7 Explanation:

Given P(m,n) ="m divides n"

Statement-A is ∃m ∀n P(m, n). Here, there exists some positive integer which divides every positive integer. It is true because there is positive integer 1 which divides every positive integer.

Statement-B is ∀n P(1, n). Here, 1 divided every positive integer. It is true.

Statement-C is ∀m ∀n P(m, n). Here, every positive integer divided every positive integer. It is false.

Statement-A is ∃m ∀n P(m, n). Here, there exists some positive integer which divides every positive integer. It is true because there is positive integer 1 which divides every positive integer.

Statement-B is ∀n P(1, n). Here, 1 divided every positive integer. It is true.

Statement-C is ∀m ∀n P(m, n). Here, every positive integer divided every positive integer. It is false.

Question 8 |

Match the following terms:

(i)(ii)(iii)(iv) | |

(ii)(iii)(i)(iv) | |

(iii)(ii)(iv)(i) | |

(iv)(iii)(ii)(i) |

Question 8 Explanation:

→ Vacuous proof is a proof that the implication p → q is true based on the fact that p is false.

→ Trivial proof is a proof that the implication p → q is true based on the fact that q is true.

→ Direct proof is a proof that the implication p → q is true that proceeds by showing that q must be true when p is true.

→ Indirect proof is a proof that the implication p → q is true that proceeds by showing that p must be false when q is false.

→ Trivial proof is a proof that the implication p → q is true based on the fact that q is true.

→ Direct proof is a proof that the implication p → q is true that proceeds by showing that q must be true when p is true.

→ Indirect proof is a proof that the implication p → q is true that proceeds by showing that p must be false when q is false.

Question 9 |

Consider the compound propositions given below as:

(a)p ∨ ~(p ∧ q)

(b)(p ∧ ~q) ∨ ~(p ∧ q)

(c)p ∧ (q ∨ r)

Which of the above propositions are tautologies?

(a)p ∨ ~(p ∧ q)

(b)(p ∧ ~q) ∨ ~(p ∧ q)

(c)p ∧ (q ∨ r)

Which of the above propositions are tautologies?

(a) and (c) | |

(b) and (c) | |

(a) and (b) | |

only (a) |

Question 9 Explanation:

Question 10 |

Which of the following property/ies a Group G must hold, in order to be an Abelian group?

(a)The distributive property

(b)The commutative property

(c)The symmetric property

(a)The distributive property

(b)The commutative property

(c)The symmetric property

(a) and (b) | |

(b) and (c) | |

(a) only | |

(b) only |

Question 10 Explanation:

An abelian group is a set, A, together with an operation • that combines any two elements a and b to form another element denoted a • b. The symbol • is a general placeholder for a
concretely given operation. To qualify as an abelian group, the set and operation, (A, •), must satisfy five requirements known as the abelian group axioms:

Closure: For all a, b in A, the result of the operation a • b is also in A.

Associativity: For all a, b and c in A, the equation (a • b) • c = a • (b • c) holds.

Identity element: There exists an element e in A, such that for all elements a in A, the equation e • a = a • e = a holds.

Inverse element: For each a in A, there exists an element b in A such that a • b = b • a = e, where e is the identity element.

Commutativity: For all a, b in A, a • b = b • a.

A group in which the group operation is not commutative is called a "non-abelian group" or "non-commutative group".

Closure: For all a, b in A, the result of the operation a • b is also in A.

Associativity: For all a, b and c in A, the equation (a • b) • c = a • (b • c) holds.

Identity element: There exists an element e in A, such that for all elements a in A, the equation e • a = a • e = a holds.

Inverse element: For each a in A, there exists an element b in A such that a • b = b • a = e, where e is the identity element.

Commutativity: For all a, b in A, a • b = b • a.

A group in which the group operation is not commutative is called a "non-abelian group" or "non-commutative group".

Question 11 |

Consider the following program :

#include

main( )

{

int i, inp;

float x, term=1, sum=0;

scanf(“%d %f ”, &inp, &x);

for(i=1; i<=inp; i++)

{

term = term * x/i;

sum = sum + term ;

}

printf(“Result = %f\n”, sum);

}

The program computes the sum of which of the following series?

#include

main( )

{

int i, inp;

float x, term=1, sum=0;

scanf(“%d %f ”, &inp, &x);

for(i=1; i<=inp; i++)

{

term = term * x/i;

sum = sum + term ;

}

printf(“Result = %f\n”, sum);

}

The program computes the sum of which of the following series?

x + x ^{2} /2 + x ^{3} /3 + x^{ 4} /4 +... | |

x + x ^{2} /2! + x^{ 3} /3! + x^{ 4} /4! +... | |

1 + x ^{2} /2 + x^{ 3} /3 + x^{ 4} /4 +... | |

1 + x ^{2} /2! + x^{ 3} /3! + x^{ 4} /4! +... |

Question 11 Explanation:

In this problem, we have to find series of evaluation.

Let x=5, inp=5

Iteration-1:

for(i=1; i<=inp; i++)

/* condition true 1<=5 */

{

term = term * x/i;

/* (1*5)/1=5. Here, 5 nothing but ‘x’ */

sum = sum + term ; /* 0+5=5 */

}

Iteration-2: Here, ‘i’ becomes 2.

for(i=2; i<=inp; i++)

/* condition true 2<=5 */

{

term = term * x/i;

/* (5*5)/2 is nothing but x 2 /i */

sum = sum + term ; /* 5+12=17 */

}

;;;

;;;

Iteration-5: Here, ‘i’ becomes 5.

for(i=5; i<=inp; i++)

/* condition true 5<=5 */

{

term = term * x/i;

/* (625*5)/5! is nothing but x

sum = sum + term ;

}

So, Option-B is correct answer.

Let x=5, inp=5

Iteration-1:

for(i=1; i<=inp; i++)

/* condition true 1<=5 */

{

term = term * x/i;

/* (1*5)/1=5. Here, 5 nothing but ‘x’ */

sum = sum + term ; /* 0+5=5 */

}

Iteration-2: Here, ‘i’ becomes 2.

for(i=2; i<=inp; i++)

/* condition true 2<=5 */

{

term = term * x/i;

/* (5*5)/2 is nothing but x 2 /i */

sum = sum + term ; /* 5+12=17 */

}

;;;

;;;

Iteration-5: Here, ‘i’ becomes 5.

for(i=5; i<=inp; i++)

/* condition true 5<=5 */

{

term = term * x/i;

/* (625*5)/5! is nothing but x

^{5} /i! */sum = sum + term ;

}

So, Option-B is correct answer.

Question 12 |

Consider the following two statements:

(a) A publicly derived class is a subtype of its base class.

(b) Inheritance provides for code reuse.

(a) A publicly derived class is a subtype of its base class.

(b) Inheritance provides for code reuse.

Both the statements (a) and (b) are correct. | |

Neither of the statements (a) and (b) are correct | |

Statement (a) is correct and (b) is incorrect | |

Statement (a) is incorrect and (b) is correct. |

Question 12 Explanation:

→ A publicly derived class is a subtype of its base class.

→ Inheritance is defined as deriving new classes (sub classes) from existing ones (super class or base class) and forming them into a hierarchy of classes.

→ Inheritance allows programmers to create classes that are built upon existing classes, to specify a new implementation while maintaining the same behaviors (realizing an interface), to reuse code and to independently extend original software via public classes and interfaces.

→ Inheritance is defined as deriving new classes (sub classes) from existing ones (super class or base class) and forming them into a hierarchy of classes.

→ Inheritance allows programmers to create classes that are built upon existing classes, to specify a new implementation while maintaining the same behaviors (realizing an interface), to reuse code and to independently extend original software via public classes and interfaces.

Question 13 |

Consider a “CUSTOMERS” database table having a column “CITY” filled with all the names of Indian cities (in capital letters). The SQL statement that finds all cities that have “GAR” somewhere in its name, is:

Select * from customers where city = ‘%GAR%’; | |

Select * from customers where city = ‘$GAR$’; | |

Select * from customers where city like ‘%GAR%’; | |

Select * from customers where city as ‘%GAR’; |

Question 13 Explanation:

In above question a specific pattern "GAR" is given for pattern matching.

In SQL "LIKE" clause is used for pattern matching. For LIKE clause we have two wild cards:

1. "%" which represents any sequence of "0" or more characters.

2. "_" is used to replace a single character.

So, Option C is the correct answer because they have used LIKE clause along with "%" which indicates any number of character can be present before and after "GAR" pattern.

In SQL "LIKE" clause is used for pattern matching. For LIKE clause we have two wild cards:

1. "%" which represents any sequence of "0" or more characters.

2. "_" is used to replace a single character.

So, Option C is the correct answer because they have used LIKE clause along with "%" which indicates any number of character can be present before and after "GAR" pattern.

Question 14 |

Match the following database terms to their function:

(iv)(iii)(i)(ii) | |

(ii)(iv)(i)(iii) | |

(ii)(iv)(iii)(i) | |

(iv)(iii)(ii)(i) |

Question 14 Explanation:

Normalization: Normalization is used to reduce the data redundancy in the database

Data Dictionary: Data dictionary contains meta describing database structure.

Referential Integrity: Referential integrity enforces match of primary key to foreign key

External Schema: Define view(s) of the database for particular user(s)

Data Dictionary: Data dictionary contains meta describing database structure.

Referential Integrity: Referential integrity enforces match of primary key to foreign key

External Schema: Define view(s) of the database for particular user(s)

Question 15 |

In general, in a recursive and non-recursive implementation of a problem (program) :

Both time and space complexities are better in recursive than in non-recursive program. | |

Both time and space complexities are better in non-recursive than in recursive program | |

Time complexity is better in recursive version but space complexity is better in non-recursive version of the program. | |

Space complexity is better in recursive version but time complexity is better in non-recursive version of the program. |

Question 15 Explanation:

Both time and space complexities are better in non-recursive than in recursive program. We can also call non-recursive programs are iterative programs. Iterative programs time and space complexity is better than recursive program.

→ Recursion uses more memory, but is sometimes clearer and more readable. Using loops increases the performance, but recursion can sometimes be better for the programmer (and his performance). Depends on what we you want to implement, and what's more important for you (readability, performance...)

Example: Factorial of a number.

→ Recursion uses more memory, but is sometimes clearer and more readable. Using loops increases the performance, but recursion can sometimes be better for the programmer (and his performance). Depends on what we you want to implement, and what's more important for you (readability, performance...)

Example: Factorial of a number.

Question 16 |

A three dimensional array in ‘C’ is declared as int A[x][y][z]. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer):

&A[0][0][0] + w(y* z* q + z* p + r) | |

&A[0][0][0] + w(y* z* p + z*q + r) | |

&A[0][0][0] + w(x* y* p + z* q+ r) | |

&A[0][0][0] + w(x* y* q + z* p + r) |

Question 17 |

In C++, which system - provided function is called when no handler is provided to deal with an exception?

terminate( ) | |

unexpected( ) | |

abort( ) | |

kill( ) |

Question 17 Explanation:

In some cases, the exception handling mechanism fails and a call to void terminate() is made. This terminate() call occurs in any of the following situations:

The exception handling mechanism cannot find a handler for a thrown exception. The following cases are more specific:

1. During stack unwinding, a destructor throws an exception and that exception is not handled.

2. The expression that is thrown also throws an exception, and that exception is not handled.

3. The constructor or destructor of a nonlocal static object throws an exception, and the exception is not handled.

4. A function registered with atexit() throws an exception, and the exception is not handled. The following demonstrates this:

A throw expression without an operand tries to rethrow an exception, and no exception is presently being handled.

The exception handling mechanism cannot find a handler for a thrown exception. The following cases are more specific:

1. During stack unwinding, a destructor throws an exception and that exception is not handled.

2. The expression that is thrown also throws an exception, and that exception is not handled.

3. The constructor or destructor of a nonlocal static object throws an exception, and the exception is not handled.

4. A function registered with atexit() throws an exception, and the exception is not handled. The following demonstrates this:

A throw expression without an operand tries to rethrow an exception, and no exception is presently being handled.

Question 18 |

Which of the following provides the best description of an entity type?

A specific concrete object with a defined set of processes (e.g. Jatin with diabetes) | |

A value given to a particular attribute (e.g. height - 230 cm) | |

A thing that we wish to collect data about zero or more, possibly real world examples of it may exist | |

A template for a group of things with the same set of characteristics that may exist in the real world |

Question 18 Explanation:

A template for a group of things with the same set of characteristics that may exist in the real world is best description of entity type.

Except option-D, remaining specifying their attributes.

Except option-D, remaining specifying their attributes.

Question 19 |

Data which improves the performance and accessibility of the database are called:

Indexes | |

User Data | |

Application Metadata | |

Data Dictionary |

Question 19 Explanation:

→ An index is a data which improves the performance and accessibility of the database. An index is a copy of selected columns of data from a table that can be searched very efficiently
that also includes a low-level disk block address or direct link to the complete row of data it was copied from.

→ To create the application metadata XML file containing details of all the Business Process Server applications that need to be packaged and published.

→ A data dictionary (or) metadata repository is a "centralized repository of information about data such as meaning, relationships to other data, origin, usage, and format".

→ To create the application metadata XML file containing details of all the Business Process Server applications that need to be packaged and published.

→ A data dictionary (or) metadata repository is a "centralized repository of information about data such as meaning, relationships to other data, origin, usage, and format".

Question 20 |

A relation R = {A, B, C, D, E, F,G} is given with following set of functional dependencies:
F = {AD → E, BE → F, B → C, AF → G}
Which of the following is a candidate key ?

A | |

AB | |

ABC | |

ABD |

Question 20 Explanation:

Since no functional dependency is having ABD in their right hand side, so every key should include ABD as a part of them to become a candidate key.

Now check whether ABD is itself a candidate key or not by finding the closure.

(ABD)

since ABD can identify each attribute of the given relation uniquely.

So, ABD is the candidate key.

Now check whether ABD is itself a candidate key or not by finding the closure.

(ABD)

^{+} = { ABDCEFG }since ABD can identify each attribute of the given relation uniquely.

So, ABD is the candidate key.

Question 21 |

Which of the following services is not provided by wireless access point in 802.11 WLAN ?

Association | |

Disassociation | |

Error correction | |

Integration |

Question 21 Explanation:

→ Wireless LANs are use high frequency radio waves instead of cables for connecting the
devices in LAN. Users connected by WLANs can move around within the area of network coverage. Most WLANs are based upon the standard IEEE 802.11 or WiFi.

→ A wireless access point (WAP), or more generally just access point (AP), is a networking hardware device that allows other Wi-Fi devices to connect to a wired network.

→ Wireless access point in 802.11 WLAN services are association, association and Integration.

→ A wireless access point (WAP), or more generally just access point (AP), is a networking hardware device that allows other Wi-Fi devices to connect to a wired network.

→ Wireless access point in 802.11 WLAN services are association, association and Integration.

Question 22 |

Which of the following fields in IPv4 datagram is not related to fragmentation?

Type of service | |

Fragment offset | |

Flags | |

Identification |

Question 22 Explanation:

→ The Identification field along with the foreign and local internet address and the protocol ID.

→ Fragment offset field along with Don't Fragment and More Fragment flags in the IP protocol header are used for fragmentation and reassembly of IP packets.

→ The More Fragment(MF) flag is set for all the fragment packets except the last one.

→ Fragment offset field along with Don't Fragment and More Fragment flags in the IP protocol header are used for fragmentation and reassembly of IP packets.

→ The More Fragment(MF) flag is set for all the fragment packets except the last one.

Question 23 |

Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is __________.

400 bps | |

800 bps | |

1600 bps | |

3200 bps |

Question 23 Explanation:

Given data,

-- 4 channels are multiplexed using TDM

-- Multiplex 1 byte per channel. So, total 4 bytes=4 channels.

-- Each channel sends 100 bytes/sec

-- Bit rate=?

Bit rate= 100 * 4 bytes(1 byte=8 bits)

= 100 * 4 *8 (we are converting into bits because we are finding bit rate)

= 3200 bps

-- 4 channels are multiplexed using TDM

-- Multiplex 1 byte per channel. So, total 4 bytes=4 channels.

-- Each channel sends 100 bytes/sec

-- Bit rate=?

Bit rate= 100 * 4 bytes(1 byte=8 bits)

= 100 * 4 *8 (we are converting into bits because we are finding bit rate)

= 3200 bps

Question 24 |

In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in adjacent cells. If 840 frequencies are available, how many can be used in a given cell?

280 | |

110 | |

140 | |

120 |

Question 24 Explanation:

-- Mobile phone system with hexagonal( equals to 6) cells

-- 840 frequencies are available.

-- How many can be used in a cell?

In this we need maximum 3 unique cells are required.

Total cells used with frequency is = 840/3

= 280 frequencies

-- 840 frequencies are available.

-- How many can be used in a cell?

In this we need maximum 3 unique cells are required.

Total cells used with frequency is = 840/3

= 280 frequencies

Question 25 |

Using p = 3, q = 13, d = 7 and e = 3 in the RSA algorithm, what is the value of ciphertext for a plain text 5 ?

8 | |

16 | |

26 | |

33 |

Question 25 Explanation:

Step-1: p=3, q=13, d=7 and e=3

n=p*q [n is the modulus for the public key and the private keys]

=39

Step-2: Calculate the totient: φ(n)= (p−1)*(q−1)

= 2*12

= 24

Step-3: Find cipher text we have an formula

C = M

= 5

= 125 mod 39

= 8

n=p*q [n is the modulus for the public key and the private keys]

=39

Step-2: Calculate the totient: φ(n)= (p−1)*(q−1)

= 2*12

= 24

Step-3: Find cipher text we have an formula

C = M

^{e} mod n= 5

^{3} mod 39= 125 mod 39

= 8

Question 26 |

A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries:
Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU ?

1024, 3072, 4096, 6144 | |

1234, 4012, 5000, 6200 | |

1020, 3012, 6120, 8100 | |

2021, 4050, 5112, 7100 |

Question 26 Explanation:

Step-1: Total number of blocks=4

Total number of pages=8

Virtual memory has a page size of 1K(1024) words.

Step-2: Page ranges

Total number of pages=8

Virtual memory has a page size of 1K(1024) words.

Step-2: Page ranges

Question 27 |

Suppose that the number of instructions executed between page fault is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further,
consider that a normal instruction takes one microsecond, but if a page fault occurs, it takes 2001 microseconds. If a program takes 60 sec to run, during which time it gets 15,000 page faults, how long would it take to run if twice as much memory were
available?

60 sec | |

30 sec | |

45 sec | |

10 sec |

Question 27 Explanation:

Normal instruction takes 1 microsecond = 10

→ Instruction with page fault takes 2001 microseconds so page fault takes additional 2000 microsecond

→ The given program takes 60 sec to run and in that program there were 15000 p.f

→ Since time taken by 1 page fault is 2000 microseconds, time taken by all 15000 page faults = 15000*2000 microseconds

=30 sec

→ Out of overall 60 sec of the program time 30 sec are taken for page faults, remaining 30 sec are consumed by program execution.

→ If memory is doubled we will have more memory available to hold the pages and hence page faults also reduce. It is given in the question that "if memory is doubled then the mean interval between the page faults is also doubled" this is nothing but "number of page faults will be reduced by half".

→ Earlier 30 sec were needed by program instructions and 30 sec for page faults.

→ Now program execution takes 30 sec and page fault will take only 15 sec.

So new total time =30+15

=45 sec

^{-6}sec→ Instruction with page fault takes 2001 microseconds so page fault takes additional 2000 microsecond

→ The given program takes 60 sec to run and in that program there were 15000 p.f

→ Since time taken by 1 page fault is 2000 microseconds, time taken by all 15000 page faults = 15000*2000 microseconds

=30 sec

→ Out of overall 60 sec of the program time 30 sec are taken for page faults, remaining 30 sec are consumed by program execution.

→ If memory is doubled we will have more memory available to hold the pages and hence page faults also reduce. It is given in the question that "if memory is doubled then the mean interval between the page faults is also doubled" this is nothing but "number of page faults will be reduced by half".

→ Earlier 30 sec were needed by program instructions and 30 sec for page faults.

→ Now program execution takes 30 sec and page fault will take only 15 sec.

So new total time =30+15

=45 sec

Question 28 |

Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. What is the time in msec to read a block of 1024 bytes from this disk?

57 sec | |

49 sec | |

48 sec | |

17 sec |

Question 28 Explanation:

Given data,

-- Bytes per track=16384

-- Rotational time=16 msec.

-- Average seek time= 40 msec

-- What is the time in msec to read a block of 1024 bytes from this disk?

Step-1: Total time= Seek Time + Rotational Latency + Transfer Time = ? + ? + 40 msec

Step-2: Transfer time = ((1024 bytes *16 ms)/ (16384 bytes )) = 1 msec

Step-3: Rotational latency = 16 msec / 2

= 8 msec

Step-4: Total time= Seek Time + Rotational Latency + Transfer Time

= 40+8+1 msec

= 49 msec

-- Bytes per track=16384

-- Rotational time=16 msec.

-- Average seek time= 40 msec

-- What is the time in msec to read a block of 1024 bytes from this disk?

Step-1: Total time= Seek Time + Rotational Latency + Transfer Time = ? + ? + 40 msec

Step-2: Transfer time = ((1024 bytes *16 ms)/ (16384 bytes )) = 1 msec

Step-3: Rotational latency = 16 msec / 2

= 8 msec

Step-4: Total time= Seek Time + Rotational Latency + Transfer Time

= 40+8+1 msec

= 49 msec

Question 29 |

A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows:
The smallest value of ‘x’ for which the above system in safe state is __________.

1 | |

3 | |

2 | |

Not safe for any value of x. |

Question 29 Explanation:

Case 1: Let’s consider x=1

→ Now need of Process D can be fulfilled. After completing its execution Process D leaves its allocated resources.

→ Now Need of any either Process can’t be fulfilled. So the system is not in safe state with x=1.

Case 2: x=2

Available = 0 0 2 1 1

→ Need of Process D can be fulfilled and after its execution it release its allocated resources.

Available

= 0 0 2 1 1

D

= 1 1 1 1 0

---------------------------------- New available = 1 1 2 2 1

→ Now Need of Process ‘C’ can be fulfilled.

Available

= 1 1 3 2 1

C

= 1 1 0 1 0

----------------------------------

New available = 2 2 3 3 1

→ Now Need of Process ‘B’ can be fulfilled. So new available resources are

Available

= 2 2 3 3 1

C

= 2 0 1 1 0

----------------------------------

New available = 4 2 4 4 1

→ But now need of Process ‘A’ can’t be fulfilled with above available resources. So the system is not in safe state with x=2.

Case 3: x=3

Available = 0 0 3 1 1

→ Now Need of Process ‘D’ can be fulfilled and after its execution it will release its allocated resources.

Available

= 0 0 3 1 1

D

= 1 1 1 1 0

----------------------------------

New available = 1 1 4 2 1

→ Now Need of Process ‘C’ can be fulfilled

Available

= 1 1 4 2 1

C

= 1 1 0 1 0

----------------------------------

New available = 2 2 4 3 1

→ Now Need of Process ‘B’ can be fulfilled

Available

= 2 2 4 3 1

B

= 2 0 1 1 0

----------------------------------

New available = 4 2 5 4 1

→ But with available resource need of Process ‘A’ can’t be fulfilled. So the system is not in safe state with x=3.

→ Hence option (D) is the correct answer.

Question 30 |

In Unix, the login prompt can be changed by changing the contents of the file __________

contrab | |

init | |

gettydefs | |

inittab |

Question 30 Explanation:

gettydefs : speed and tty settings used by getty. The file /etc/gettydefs contains information used by getty(1m) to set up the speed and tty settings for a line. It supplies information on what the login-prompt should look like. It also supplies the speed to try next if the user indicates the current speed is not correct by typing a character.

init (short for initialization) is the first process started during booting of the computer system. Init is a daemon process that continues running until the system is shut down.

inittab - format of the inittab file used by the sysv-compatible init process. The inittab file describes which processes are started at bootup and during normal operation (e.g. /etc/init.d/boot, /etc/init.d/rc, gettys...). Init(8) distinguishes multiple runlevels, each of which can have its own set of processes that are started. Valid runlevels are 0-6 plus A, B, and C for on demand entrie.

crontab (short for "cron table") is a list of commands that are scheduled to run at regular time intervals on your computer system. The crontab command opens the crontab for editing, and lets you add, remove, or modify scheduled tasks.

init (short for initialization) is the first process started during booting of the computer system. Init is a daemon process that continues running until the system is shut down.

inittab - format of the inittab file used by the sysv-compatible init process. The inittab file describes which processes are started at bootup and during normal operation (e.g. /etc/init.d/boot, /etc/init.d/rc, gettys...). Init(8) distinguishes multiple runlevels, each of which can have its own set of processes that are started. Valid runlevels are 0-6 plus A, B, and C for on demand entrie.

crontab (short for "cron table") is a list of commands that are scheduled to run at regular time intervals on your computer system. The crontab command opens the crontab for editing, and lets you add, remove, or modify scheduled tasks.

Question 31 |

A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?

p ^{n} | |

p | |

(2 ^{n} - 1)p+1 | |

(p + 1) ^{n} |

Question 31 Explanation:

Option-A: This is the maximum number of distinct tuples that you can form with p distinct values per dimensions.

Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.

Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.

Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.

Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.

Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.

Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.

Question 32 |

Suppose that from given statistics, it is known that meningitis causes stiff neck 50% of the time, that the proportion of persons having meningitis is 1 / 50000 , and that the proportion of people having stiff neck is 1 / 20. Then the percentage of people who had meningitis and complain about stiff neck is:

0.01% | |

0.02% | |

0.04% | |

0.05% |

Question 33 |

__________ system is market oriented and is used for data analysis by knowledge workers including Managers, Executives and Analysts.

OLTP | |

OLAP | |

Data System | |

Market System |

Question 33 Explanation:

→ OLAP system is market oriented and is used for data analysis by knowledge workers including Managers, Executives and Analysts.

→ OLAP include business reporting for sales, marketing, management reporting, business process management (BPM), budgeting and forecasting, financial reporting and similar areas, with new applications emerging, such as agriculture.

→ OLAP include business reporting for sales, marketing, management reporting, business process management (BPM), budgeting and forecasting, financial reporting and similar areas, with new applications emerging, such as agriculture.

Question 34 |

__________ allows selection of the relevant information necessary for the data warehouse.

The Top - Down View | |

Data Warehouse View | |

Datasource View | |

Business Query View |

Question 34 Explanation:

→ Top-Down is also known as Inmon’s Top Down Approach, data warehouse is built first. Inmon defines a data warehouse as a centralized repository for the entire enterprise. Dimensional data marts are created only after the complete data warehouse has been created.

→ Bottom-Up is also called as Kimball’s bottom up approach, the most important business aspects or departments, data marts are created first. A data mart provide a thin view into the organisational data and addresses a single business area. These data marts are then integrated into larger data warehouse to build a complete data warehouse. The integration of data marts is implemented using Kimball’s data warehousing architecture which is also known as data warehouse bus (BUS).

→ Bottom-Up is also called as Kimball’s bottom up approach, the most important business aspects or departments, data marts are created first. A data mart provide a thin view into the organisational data and addresses a single business area. These data marts are then integrated into larger data warehouse to build a complete data warehouse. The integration of data marts is implemented using Kimball’s data warehousing architecture which is also known as data warehouse bus (BUS).

Question 35 |

The hash function used in double hashing is of the form:

h (k, i) = (h _{1} (k) + h_{ 2} (k) + i) mod m | |

h (k, i) = (h _{1} (k) + h_{ 2} (k) - i) mod m | |

h (k, i) = (h _{1} (k) + i h_{ 2} (k)) mod m | |

h (k, i) = (h _{1} (k) - i h_{ 2} (k)) mod m |

Question 35 Explanation:

Double Hashing with second hash function is h(k,i) = (h

h

h

First try h(k,0) = h 1 (k), if it is occupied, try h(k,1) etc..,

Advantage: less clusters, uses Θ(m*m) permutations of index addressing sequences.

_{1} (k) + i·h_{ 2} (k)) mod m 0≤ i ≤m-1h

_{1} → hash functionh

_{2} → Step functionFirst try h(k,0) = h 1 (k), if it is occupied, try h(k,1) etc..,

Advantage: less clusters, uses Θ(m*m) permutations of index addressing sequences.

Question 36 |

In the following graph, discovery time stamps and finishing time stamps of Depth First Search (DFS) are shown as x/y, where x is discovery time stamp and y is finishing time stamp.

It shows which of the following depth first forest?

It shows which of the following depth first forest?

{a, b, e} {c, d, f, g, h} | |

{a, b, e} {c, d, h} {f, g} | |

{a, b, e} {f, g} {c, d} {h} | |

{a, b, c, d} {e, f, g} {h} |

Question 36 Explanation:

A DFS starting at some vertex ‘v’ explores the graph by building up a tree that contains all vertices that are reachable from ‘v’ and all edges that are used to reach these vertices. We call this tree a DFS tree. A complete DFS exploring the full graph (and not only the part
reachable from a given vertex ‘v’) builds up a collection of trees, or forest, called a DFS forest.

Based on definition of DFS forest, option-A is correct.

Based on definition of DFS forest, option-A is correct.

Question 37 |

The number of disk pages access in B-Tree search, where ‘h’ is height, ‘n’ is the number of keys, and ‘t’ is the minimum degree, is:

θ(log _{n} h*t) | |

θ(log _{t} n*h) | |

θ(log _{h} n) | |

θ(log _{t} n) |

Question 37 Explanation:

The number of disk pages access in B-Tree search, where ‘h’ is height, ‘n’ is the number of keys, and ‘t’ is the minimum degree, is θ(log

Note: B-Tree search operation best,average and worst case will take θ(logn).

_{t} n).Note: B-Tree search operation best,average and worst case will take θ(logn).

Question 38 |

The inorder traversal of the following tree is:

2 3 4 6 7 13 15 17 18 18 20 | |

20 18 18 17 15 13 7 6 4 3 2 | |

15 13 20 4 7 1718 2 3 6 18 | |

2 4 3 13 7 6 15 17 20 18 18 |

Question 38 Explanation:

Inorder traversal through Left, Root and Right. Inorder yield procedure given below.

Question 39 |

An ideal sort is an in-place-sort whose additional space requirement is __________.

O (log _{2} n) | |

O (n log _{2} n) | |

O (1) | |

O (n) |

Question 39 Explanation:

The ideal sorting algorithm would have the following properties:

1. Stable: Equal keys aren't reordered.

2. Operates in place, requiring O(1) extra space.

3. Worst case O(nlog(n)) key comparisons.

4. Worst case O(n) swaps.

5. Adaptive: Speeds up to O(n) when data is nearly sorted or when there are few unique keys.

1. Stable: Equal keys aren't reordered.

2. Operates in place, requiring O(1) extra space.

3. Worst case O(nlog(n)) key comparisons.

4. Worst case O(n) swaps.

5. Adaptive: Speeds up to O(n) when data is nearly sorted or when there are few unique keys.

Question 40 |

Which of the following is not a congestion policy at network layer?

Flow Control Policy | |

Packet Discard Policy | |

Packet Lifetime Management Policy | |

Routing Algorithm |

Question 40 Explanation:

Network Layer congestion policies:

1.Preallocation of Resources

2.Traffic Shaping

3.Discarding Packets (No Preallocation)

4.Isarithmic Congestion Control

5.Virtual Circuits Admission Control

6.Choke Packets

7.Packet Discard

8. Packet Lifetime Management

9. Routing Algorithm

Transport layer congestion policies:

1. Flow Control, etc..,

1.Preallocation of Resources

2.Traffic Shaping

3.Discarding Packets (No Preallocation)

4.Isarithmic Congestion Control

5.Virtual Circuits Admission Control

6.Choke Packets

7.Packet Discard

8. Packet Lifetime Management

9. Routing Algorithm

Transport layer congestion policies:

1. Flow Control, etc..,

Question 41 |

Loop unrolling is a code optimization technique:

that avoids tests at every iteration of the loop. | |

that improves performance by decreasing the number of instructions in a basic block. | |

that exchanges inner loops with outer loops | |

that reorders operations to allow multiple computations to happen in parallel |

Question 41 Explanation:

→ Loop unrolling is a code optimization technique that avoids tests at every iteration of the loop.

→ The goal of loop unwinding is to increase a program's speed by reducing (or) eliminating instructions that control the loop, such as pointer arithmetic and "end of loop" tests on each iteration.

→ The goal of loop unwinding is to increase a program's speed by reducing (or) eliminating instructions that control the loop, such as pointer arithmetic and "end of loop" tests on each iteration.

Question 42 |

What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions?

mov al, 15

mov ah, 15

xor al, al

mov cl, 3

shr ax, cl

mov al, 15

mov ah, 15

xor al, al

mov cl, 3

shr ax, cl

0F00 h | |

0F0F h | |

01E0 h | |

FFFF h |

Question 42 Explanation:

Step-1: mov al, 15 ← Move 15 to lower part of 'ax' register

Step-2: mov ah, 15 ← Move 15 to higher part of 'ax' register

Step-3: xor al,al ← Perform Xor operation.

convert 15 into binary number is 00001111

00001111 ← al

00000000 ← al

-------------

00000000

--------------

Step-4: move cl,3 ← Move 3 to lower part of 'cx' register

Step-5: shr ax, cl ← It will shift right and rotate content of 'ax' register

Before shifting ‘ax’ value is 0000 1111 0000 0000

After 3 shift right operations then ‘ax’ will become 0000 0001 1110 0000

The value is nothing but 01E0 h (or) (01E0)

Step-2: mov ah, 15 ← Move 15 to higher part of 'ax' register

Step-3: xor al,al ← Perform Xor operation.

convert 15 into binary number is 00001111

00001111 ← al

00000000 ← al

-------------

00000000

--------------

Step-4: move cl,3 ← Move 3 to lower part of 'cx' register

Step-5: shr ax, cl ← It will shift right and rotate content of 'ax' register

Before shifting ‘ax’ value is 0000 1111 0000 0000

After 3 shift right operations then ‘ax’ will become 0000 0001 1110 0000

The value is nothing but 01E0 h (or) (01E0)

_{ 16}Question 43 |

Which of the following statements is false?

Top-down parsers are LL parsers where first L stands for left - to - right scan and second L stands for a leftmost derivation. | |

(000)* is a regular expression that matches only strings containing an odd number of zeroes, including the empty string. | |

Bottom-up parsers are in the LR family, where L stands for left - to - right scan and R stands for rightmost derivation. | |

The class of context - free languages is closed under reversal. That is, if L is any context- free language, then the language L R = {w R : w∈L} is context - free. |

Question 43 Explanation:

TRUE: Top-down parsers are LL parsers where first L stands for left - to - right scan and second L stands for a leftmost derivation.

FALSE: (000)* is a regular expression that matches any string containing at least 3 number of zeroes, including the empty string.

TRUE: Bottom-up parsers are in the LR family, where L stands for left - to - right scan and R stands for rightmost derivation.

TRUE: The class of context - free languages is closed under reversal. That is, if L is any context - free language, then the language L

FALSE: (000)* is a regular expression that matches any string containing at least 3 number of zeroes, including the empty string.

TRUE: Bottom-up parsers are in the LR family, where L stands for left - to - right scan and R stands for rightmost derivation.

TRUE: The class of context - free languages is closed under reversal. That is, if L is any context - free language, then the language L

^{R} = {w^{ R} : w∈L} is context - free.Question 44 |

System calls are usually invoked by using :

A privileged instruction | |

An indirect jump | |

A software interrupt | |

Polling |

Question 44 Explanation:

→ System calls are usually invoked by using a software interrupt.

→ Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.

→ Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.

Question 45 |

The __________ transfers the executable image of a C++ program from hard disk to main memory.

Compiler | |

Linker | |

Debugger | |

Loader |

Question 45 Explanation:

→ The loader transfers the executable image of a C++ program from hard disk to main memory.

→ Loader is the program of the operating system which loads the executable from the disk into the primary memory(RAM) for execution. It allocates the memory space to the executable module in main memory and then transfers control to the beginning instruction of the program .

→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running. Once loading is complete, the operating system starts the program by passing control to the loaded program code.

→ Loader is the program of the operating system which loads the executable from the disk into the primary memory(RAM) for execution. It allocates the memory space to the executable module in main memory and then transfers control to the beginning instruction of the program .

→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running. Once loading is complete, the operating system starts the program by passing control to the loaded program code.

Question 46 |

In software testing, how the error, fault and failure are related to each other?

Error leads to failure but fault is not related to error and failure. | |

Fault leads to failure but error is not related to fault and failure. | |

Error leads to fault and fault leads to failure. | |

Fault leads to error and error leads to failure. |

Question 46 Explanation:

→ Error: A human action that produces an incorrect result. This is also sometimes referred to as Mistake.

→ Fault: A manifestation of an error in software, also known as Defect or Bug.

→ Failure: A deviation of the software from its expected delivery or service.

Relation between Error,Fault and Failure

→ Fault: A manifestation of an error in software, also known as Defect or Bug.

→ Failure: A deviation of the software from its expected delivery or service.

Relation between Error,Fault and Failure

Question 47 |

Which of the following is not a software process model?

Prototyping | |

Iterative | |

Time-boxing | |

Glass-boxing |

Question 47 Explanation:

Software process models of Prototype and Iterative we know before also.

Time boxing model:

The development is done iteratively as in the iterative enhancement model and each iteration is done in a timebox of fixed duration. The functionality to be developed is adjusted to fit the duration of the timebox. Moreover, each timebox is divided into a sequence of fixed stages where each stage performs a clearly defined task (analysis, implementation, and deploy) that can be done independently. This model also requires that the time duration of each stage is approximately equal so that pipelining concept is employed to have the reduction in development time and product releases.

Time boxing model:

The development is done iteratively as in the iterative enhancement model and each iteration is done in a timebox of fixed duration. The functionality to be developed is adjusted to fit the duration of the timebox. Moreover, each timebox is divided into a sequence of fixed stages where each stage performs a clearly defined task (analysis, implementation, and deploy) that can be done independently. This model also requires that the time duration of each stage is approximately equal so that pipelining concept is employed to have the reduction in development time and product releases.

Question 48 |

How many solutions are there for the equation x + y + z + u = 29 subject to the constraints that x ≥ 1, y ≥ 2, z ≥ 3 and u ≥ 0?

4960 | |

2600 | |

23751 | |

8855 |

Question 48 Explanation:

Given data,

-- The equation x+y+z+u = 29

-- Constraints are x ≥ 1, y ≥ 2, z ≥ 3 and u ≥ 0

-- Possible solutions?

Step-1: Combine all constraints (1+2+3+0) =6

Step-2: Subtract all constraints values with total equation number 29-6 =23.

Possibility to get repetition of a numbers of x,y and z but no chance for ‘u’ because its value is 0.

Step-3: So, Subtract repeated values into total equation value

= 29-3

= 26

Step-4: Possible solutions=

= 2600

-- The equation x+y+z+u = 29

-- Constraints are x ≥ 1, y ≥ 2, z ≥ 3 and u ≥ 0

-- Possible solutions?

Step-1: Combine all constraints (1+2+3+0) =6

Step-2: Subtract all constraints values with total equation number 29-6 =23.

Possibility to get repetition of a numbers of x,y and z but no chance for ‘u’ because its value is 0.

Step-3: So, Subtract repeated values into total equation value

= 29-3

= 26

Step-4: Possible solutions=

^{26}C_{ 23}= 2600

Question 49 |

A unix file system has 1-KB blocks and 4-byte disk addresses. What is the maximum file size if i-nodes contain 10 direct entries and one single, double and triple indirect entry each?

32 GB | |

64 GB | |

16 GB | |

1 GB |

Question 49 Explanation:

Block size = 1KB

Size of one address = 2

No. of addresses a block can contain/point =2

Max. file size =(10+2

=16 GB

Size of one address = 2

^{2}ByteNo. of addresses a block can contain/point =2

^{10}/ 2^{2}=2^{8}Max. file size =(10+2

^{8}+(2^{8}* 2^{8})+(2^{8}* 2^{8}* 2^{8}))2^{10}=16 GB

Question 50 |

__________ uses electronic means to transfer funds directly from one account to another rather than by cheque or cash.

M-Banking | |

E-Banking | |

O-Banking | |

C-Banking |

Question 50 Explanation:

E-Banking uses electronic means to transfer funds directly from one account to another rather than by cheque or cash.

E-banking (or) Online banking (or) internet banking:

It is an electronic payment system that enables customers of a bank or other financial institution to conduct a range of financial transactions through the financial institution's website. The online banking system will typically connect to or be part of the core banking system operated by a bank and is in contrast to branch banking which was the traditional way customers accessed banking services.

E-banking (or) Online banking (or) internet banking:

It is an electronic payment system that enables customers of a bank or other financial institution to conduct a range of financial transactions through the financial institution's website. The online banking system will typically connect to or be part of the core banking system operated by a bank and is in contrast to branch banking which was the traditional way customers accessed banking services.

There are 50 questions to complete.